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adkikani
If x is an integer, what is the value of x ?

(1) \(\frac{1}{5}\) < \(\frac{1}{(x+1)}\) < \(\frac{1}{2}\)

(2) \((x - 3)(x - 4)\) = 0
(1) 1/(x+1) can be 1/3 or 1/4
therefore, x can be 2 or 3. Insufficient

(2) x = 3 or 4. insufficient

(1)+(2), x = 3. Sufficient

C is correct.­
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adkikani
If x is an integer, what is the value of x ?

(1) \(\frac{1}{5}\) < \(\frac{1}{(x+1)}\) < \(\frac{1}{2}\)

(2) \((x - 3)(x - 4)\) = 0

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.

Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)

1/5 < 1/(x+1) < 1/2
⇔ 5 > x+1 > 2
⇔ 4 ≥ x+1 ≥ 3
⇔ 3 ≥ x ≥ 2
⇔ x = 2 or x = 3.

Since condition 1) does not yield a unique solution, it is not sufficient

Condition 2)
(X-3)(x-4) = 0
⇔ x = 3 or x = 4

Since condition 2) does not yield a unique solution, it is not sufficient

Conditions 1) & 2)
We have x = 2 or x = 3 from condition 1).
We have x = 3 or x = 4 from condition 2).
When we consider both conditions together, we have x = 3 only.

Since both conditions together yield a unique solution, they are sufficient

Therefore, C is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.­
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VeritasKarishma Bunuel GMATBusters

Conceptual question regarding Statement 1 :

If e.g we were only told that the expression is less than 0.5 as opposed to the range that we have been given now i.e the expression lies between 0.2 and 0.5, then in that case would we have been able to cross multiply the expression x + 1 as we do not know whether the expression is positive or not. Now we know that the expression is positive and hence we can find the values of x. Am I right in concluding that in the hypothetical case that I just presented we would not have been able to find any value of x because of sign ambiguity ?

Will appreciate a response please.
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altairahmad
VeritasKarishma Bunuel GMATBusters

Conceptual question regarding Statement 1 :

If e.g we were only told that the expression is less than 0.5 as opposed to the range that we have been given now i.e the expression lies between 0.2 and 0.5, then in that case would we have been able to cross multiply the expression x + 1 as we do not know whether the expression is positive or not. Now we know that the expression is positive and hence we can find the values of x. Am I right in concluding that in the hypothetical case that I just presented we would not have been able to find any value of x because of sign ambiguity ?

Will appreciate a response please.

1/(x + 1) < 1/2

1/(x + 1) - 1/2 < 0

[2 - (x + 1)] / 2(x + 1) < 0

(x - 1)/2(x + 1) > 0 (multiplied both sides by -1)

Transition points are -1 and 1.
x > 1 or x < -1 (for the expression to be positive)
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adkikani
If x is an integer, what is the value of x ?

(1) \(\frac{1}{5}\) < \(\frac{1}{(x+1)}\) < \(\frac{1}{2}\)

(2) \((x - 3)(x - 4)\) = 0
(1) Taking reciprocals all over we get 5 > x+1 > 2 which means x+1 can either be 3 or 4 which means x can either be 2 or 3. Insufficient.

(2) x = 3 or x = 4. Insufficient.

Combining the two we get:
x = 3. Sufficient.

Hence, C.­
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